Question

(25 pts.) Let the random variable X have pdf f(x) = { 0<x<1 1<isa Generate a random variable from f(x) using (a) The inverse-transform method (b) The accept-reject method, using the proposal density 9(x) = 0sos

Answer 1

a)Inverse Transform method:

If $\fn_jvn U\sim Unif\left ( 0,1 \right )$ has distribution, then $\fn_jvn X$ such that $\fn_jvn X=F^{-1}_X\left ( U \right )$ is an observation from the probability distribution $\fn_jvn F_X\left ( X \right )$ , this means that we can generate observations from the distribution $\fn_jvn F_X\left ( X \right )$ by generating $\fn_jvn U\left ( 0,1 \right )$ random variables (which most software programs can do easily) and applying the $\fn_jvn F^{-1}_X$ transformation.

Here

F(x) = 0<x< 1 10.5x – 0.25 ,1<x<5/2 4

The inverse function is

.

2V0 0<U< 0.25 F'(U) = { 20+ 0.5 ,0.25<U<1

To generate one random variable from F, generate , then

U = 0.3546786

X = 210.3546786) +0.5 = 1.2093572

b) Acceptance-Rejection Algorithm for continuous random variables

1. Generate a RV distributed as .

2. Generate (independent from )

3. If
Us A (Y) 0<9×(y)
, then set
X = Y(accept)
; otherwise go back to 1, (“reject”).

Take
Y~110,52)
. Then
fy (y) = 0.4; 0 <y <2.5

To find the maximum value of ,
gx (x) - ** f (x) = 257
,

Which is maximum when
X = Xmax = 2.5

So,
c = 9x (2.5) - (2.5) = 2 fy (2.5) 0.4
.

Iterations for generating a random number from G.

Let
U = 0.3546786
and
Y = 0.8646336
.

. Reject

gx (Y) cfy (Y) (0.8646336) = 0.34585344 < 0.3546786 0.4 x 2

Let
U = 0.4317008
and
Y = 1.764366
.

gx (Y) cfy (Y) = (1.764366) L = 0.7057464 0.4 x 2

Us A (Y) 0<9×(y)
, so accept
X = 0.8646336